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weqwewe [10]
3 years ago
10

Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy

Physics
1 answer:
Anestetic [448]3 years ago
4 0

Answer:

To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Explanation:

The general formula of kinetic energy is given as follows:

K.E = \frac{1}{2}mv^{2}

where,

K.E = Kinetic Energy

m = mass of the object

v = speed of the object

So, for the marble and rock to have same kinetic energy, we can write:

K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} =  \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}

<u>Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.</u>

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Dr. Glover's office has one vendor for their practice management software and another for their electronic health record, but th
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Answer:

Interface

Explanation:

This is a classic example of Interface technology.

An interface allows different software packages to communicate without re-entering data.

Here in this case also systems are able to communicate with one another without duplicating data entry. For example, practice management software and another for their electronic health record.

5 0
3 years ago
When a light bulb is connected to a 4.5 V battery, a current of 0.16 A passes through the bulb filament. What is the resistance
larisa [96]

Answer:

R = 28.125 ohms

Explanation:

Given that,

The voltage of a bulb, V = 4.5 V

Current, I = 0.16 A

We need to find the resistance of the filament. Using Ohm's law,

V = IR

Where

R is the resistance of the filament

So,

R=\dfrac{V}{I}\\\\R=\dfrac{4.5}{0.16}\\\\R=28.125\ \Omega

So, the resistance of the filament is equal to 28.125 ohms.

5 0
3 years ago
A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine
Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = \frac{4}{3}\pi(\frac{d}{2})^3

or

Volume of the balloon = \frac{4}{3}\pi(\frac{0.296}{2})^3

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314  kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = \frac{3}{2}\times K_BT

where,

Boltzmann constant, K_B=1.3807\times10^{-23}J/K

Average kinetic energy = \frac{3}{2}\times1.3807\times10^{-23}\times292

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = \frac{3RT}{m}

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = \frac{3\times8.314\times292}{0.004}

or

rms speed = 1349.35 m/s

5 0
3 years ago
One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at
Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature(T_i)=300 K

P_i=10 atm

P_f=2 atm

Work done in iso-thermal process=P_iV_iln\frac{P_i}{P_f}

P_i=initial pressure

P_f=Final Pressure

W=10\times 2.463\times ln\frac{10}{2}=39.64 J

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is=P\Delta V

V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L

V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L

\Delta W=1\times (12.315-2.463)=9.852 J

\Delta q=\Delta W=9.852 J

\Delta U=0

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